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A) \(\mathsf{sin\,\theta=-\,\dfrac{4}{5}\qquad\qquad\theta \in 4th~quadrant.}\)
\(\mathsf{5\,sin\,\theta=-4\qquad\quad(i)}\)
• Finding \(\mathsf{cos\,\theta:}\)
Square both sides of \(\mathsf{(i):}\)
\(\mathsf{(5\,sin^2\,\theta)=4^2}\\\\
\mathsf{5^2\,sin^2\,\theta=4^2}\\\\
\mathsf{25\,sin^2\,\theta=16\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\
\mathsf{25\cdot (1-cos^2\,\theta)=16}\\\\
\mathsf{25-25\,cos^2\,\theta=16}\)
\(\mathsf{25-16=25\,cos^2\,\theta}\\\\
\mathsf{9=25\,cos^2\,\theta}\\\\
\mathsf{cos^2\,\theta=\dfrac{9}{25}}\\\\\\
\mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{cos\,\theta=\pm\,\dfrac{3}{5}}\)
But \(\mathsf{cos\,\theta}\) is also negative, because \(\mathsf{\theta}\) lies in the 4th quadrant. So,
\(\mathsf{cos\,\theta=-\,\dfrac{3}{5}\qquad\quad\checkmark}\)
• Finding \(\mathsf{tan\,\theta:}\)
\(\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\
\mathsf{tan\,\theta=\dfrac{-\,\frac{4}{5}}{-\,\frac{3}{5}}}\\\\\\
\mathsf{tan\,\theta=-\,\dfrac{4}{\diagup\hspace{-6}5}\cdot \left(\!-\,\dfrac{\diagup\hspace{-6}5}{3}\right)}\\\\\\
\mathsf{tan\,\theta=\dfrac{4}{3}\qquad\quad\checkmark}\)
• Finding \(\mathsf{cot\,\theta:}\)
\(\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\
\mathsf{cot\,\theta=\dfrac{1}{~\frac{4}{3}~}}\\\\\\
\mathsf{cot\,\theta=\dfrac{3}{4}\qquad\quad\checkmark}\)
• Finding \(\mathsf{sec\,\theta:}\)
\(\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\
\mathsf{sec\,\theta=\dfrac{1}{-\,\frac{3}{5}}}\\\\\\
\mathsf{sec\,\theta=-\,\dfrac{5}{3}\qquad\quad\checkmark}\)
• Finding \(\mathsf{csc\,\theta:}\)
\(\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\
\mathsf{csc\,\theta=\dfrac{1}{-\,\frac{4}{5}}}\\\\\\
\mathsf{csc\,\theta=-\,\dfrac{5}{4}\qquad\quad\checkmark}\)
________
B) \(\mathsf{tan\,\theta=2\qquad\qquad\theta \in 1st~quadrant.}\)
\(\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=2}\\\\\\
\mathsf{sin\,\theta=2\,cos\,\theta\qquad\quad(ii)}\)
• Finding \(\mathsf{sin\,\theta:}\)
Square both sides of \(\mathsf{(ii):}\)
\(\mathsf{(sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\
\mathsf{sin^2\,\theta=2^2\,cos^2\,\theta}\\\\
\mathsf{sin^2\,\theta=4\,cos^2\,\theta\qquad\qquad(but~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\
\mathsf{sin^2\,\theta=4-4\,sin^2\,\theta}\)
\(\mathsf{sin^2\,\theta+4\,sin^2\,\theta=4}\\\\
\mathsf{5\,sin^2\,\theta=4}\\\\
\mathsf{sin^2\,\theta=\dfrac{4}{5}}\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{4}{5}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{2}{\sqrt{5}}}\)
But sine is positive because \(\theta\) is a 1st quadrant angle:
\(\mathsf{sin\,\theta=\dfrac{2}{\sqrt{5}}\qquad\quad\checkmark}\)
• Finding \(\mathsf{cos\,\theta:}\)
\(\mathsf{sin\,\theta=2\,cos\,\theta}\\\\
\mathsf{cos\,\theta=\dfrac{1}{2}\,sin\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{1}{\diagup\hspace{-6}2}\cdot \dfrac{\diagup\hspace{-6}2}{\sqrt{5}}}\\\\\\
\mathsf{cos\,\theta=\dfrac{1}{\sqrt{5}}\qquad\quad\checkmark}\)
• Finding \(\mathsf{cot\,\theta:}\)
\(\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\
\mathsf{cot\,\theta=\dfrac{1}{2}}\qquad\quad\checkmark}\)
• Finding \(\mathsf{sec\,\theta:}\)
\(\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{~\frac{1}{\sqrt{5}}~}}\\\\\\ \mathsf{sec\,\theta=\sqrt{5}\qquad\quad\checkmark}\)
• Finding \(\mathsf{csc\,\theta:}\)
\(\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\
\mathsf{csc\,\theta=\dfrac{1}{~\frac{2}{\sqrt{5}}~}}\\\\\\
\mathsf{csc\,\theta=\dfrac{\sqrt{5}}{2}\qquad\quad\checkmark}\)
I hope this helps. =)
Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc trigonometry
